C pointer to an array
You can skip this chapter and then wait to understand the concept of C pointers, come to learn in this chapter.
If you understand the concept of pointers in C language, then you can begin to learn in this chapter. Array name is a constant pointer pointing to the first element of the array. Therefore, in the following statement:
double balance[50];
balance is a point & balance [0] pointer that address the first element of the array of the balance.Therefore, the following program fragmentp assigned to the first element of balancein the address:
double *p; double balance[10]; p = balance;
Use as a constant pointer array name is legitimate, and vice versa. Therefore, * (balance + 4) is a balance [4] legitimate way to access the data.
Once you have the address stored in the first element of p, you can use * p, * (p + 1), * (p + 2) and so to access the array elements. The following example demonstrates these concepts discussed above to:
#include <stdio.h>
int main ()
{
/* 带有 5 个元素的整型数组 */
double balance[5] = {1000.0, 2.0, 3.4, 17.0, 50.0};
double *p;
int i;
p = balance;
/* 输出数组中每个元素的值 */
printf( "使用指针的数组值\n");
for ( i = 0; i < 5; i++ )
{
printf("*(p + %d) : %f\n", i, *(p + i) );
}
printf( "使用 balance 作为地址的数组值\n");
for ( i = 0; i < 5; i++ )
{
printf("*(balance + %d) : %f\n", i, *(balance + i) );
}
return 0;
}
When the above code is compiled and executed, it produces the following results:
Using the pointer array value * (p + 0): 1000.000000 * (P + 1): 2.000000 * (P + 2): 3.400000 * (P + 3): 17.000000 * (P + 4): 50.000000 Use balance value as an array address * (balance + 0): 1000.000000 * (Balance + 1): 2.000000 * (Balance + 2): 3.400000 * (Balance + 3): 17.000000 * (Balance + 4): 50.000000
In the example above, p is a pointer to a pointer of type double, which means that it can be stored in a variable of type double. Once we have the addressp, * p value of p will be given the corresponding memory address, as demonstrated in the above examples.